Question

The range of the function, $f\left(x\right)={\cot }^{-1}(2x-x^2)$ is

• A. $[\frac{\pi }{2},\pi )$
• B. $[\frac{\pi }{4},\pi ]$
• C. $[\frac{\pi }{4},\pi )$
• D. $[0,\frac{\pi }{4}]$

Answer 1

The correct option is C $[\frac{\pi }{4},\pi )$

$y=co{t}^{-1}(2x-x^2)$

$g\left(x\right)=2x-x^2=-(x^2-2x)$

$g\left(x\right)=-(x^2-2x-1+1)=-\left(\right(x-1{)}^2-1)$

$g\left(x\right)=1-(x-1{)}^2$

Here domain is $xϵR$

Minimum of $g\left(x\right)$ is $\infty$

Maximum of $g\left(x\right)$ is $(-\infty ,1]$

$y=co{t}^{-1}\left(g\right(x\left)\right)=co{t}^{-1}\left(t\right)$

$tϵ(-\infty ,1]$